Answer:
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Explanation:
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A(n) 60.3 g ball is dropped from a height of 53.7 cm above a spring of negligible mass. The ball compresses the spring to a maxi
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In order to determine the required force to stop the car, proceed as follow:
Calculate the deceleration of the car, by using the following formula:
where,
v: final speed = 0m/s (the car stops)
vo: initial speed = 36m/s
x: distance traveled = 980m
a: deceleration of the car= ?
Solve the equation above for a, replace the values of the other parameters and simplify:
Next, consider that the formula for the force is:
where,
m: mass of the car = 820 kg
a: deceleration of the car = 0.66m/s^2
Replace the previous values and simplify:
Hence, the required force to stop the car is 542.20N
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m